How to make a 3D object array in JavaScript?

Question

lyricsInp = document.getElementById("lyrics").value;
var lines = lyricsInp.split("\n");
for (i = 0; i < lines.length; i++) {
    holder[0][i] = new Array();
    words = lines[i].replace(/[ \t\r]+/g, "###").split("###");
    for (j = 0; j < words.length; j++) {
        holder[0][i][j].word = words[j];
        holder[0][i][j].startT = 0;
        holder[0][i][j].endT = 0;
    }
}

Here I need each holder element to keep word, startT and endT, but this does not work. How do I make this happen.

What is holder[0]? What does not work? Code seems fine (unless some uneccessary global variables) — Bergi
– Bergi, Commented Nov 1, 2012 at 15:13
Try ['word'] instead of .word ... or use a javascript object instead of an array. — Chase
– Chase, Commented Nov 1, 2012 at 15:14

David G · Accepted Answer · 2012-11-01 15:14:46Z

1

Declare holder[0][i][j] as an object literal:

for (j = 0; j < words.length; j++) {
    holder[0][i][j] = {};

    holder[0][i][j].word = words[j];
    holder[0][i][j].startT = 0;
    holder[0][i][j].endT = 0;
}

answered Nov 1, 2012 at 15:14

David G

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Comments

Bergi · Accepted Answer · 2012-11-01 15:17:59Z

0

var holder = [[]];

var lyricsInp = document.getElementById("lyrics").value,
    lines = lyricsInp.split("\n");
for (var i = 0; i < lines.length; i++) {
    holder[0][i] = [];
    var words = lines[i].replace(/[ \t\r]+/g, "###").split("###");
    for (var j = 0; j < words.length; j++) {
        holder[0][i][j] = {
            word: words[j],
            startT: 0,
            endT: 0
        };
    }
}

answered Nov 1, 2012 at 15:17

Bergi

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Comments

Pavel Anossov · Accepted Answer · 2012-11-01 15:20:32Z

0

lyricsInp =  document.getElementById("lyrics").value;
var lines = lyricsInp.split("\n");
for (i=0;i<lines.length;i++){
    holder[0][i] = new Array();
    words = lines[i].replace(/[ \t\r]+/g,"###").split("###");
    for (j=0;j<words.length;j++){
        holder[0][i][j] = {
                word: words[j],
                startT: 0,
                endT: 0
            };
    }
}

edited Nov 1, 2012 at 15:20

answered Nov 1, 2012 at 15:15

Pavel Anossov

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